I've seen multiple examples of how to solve for $\sin(z) = 0$, however how may I go about solving $\sin(\sin(z))=0$?
My thoughts are to start with rewriting $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$ And then trying to find when $\sin(\frac{e^{iz}-e^{-iz}}{2i})=0$
Since sin doesn't have a well-defined inverse I'm not sure as to which method I should use to solve this. All tips appreciated
By definition
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
so that
$$\sin z=0\iff e^{2iz}=1\iff z=k\pi\;,\;\;k\in\Bbb Z$$
thus we need to solve
$$\sin z=k\pi\implies\frac{e^{iz}-e^{-iz}}{2i}=k\pi\;,\;\;k\in\Bbb Z$$
let $x=e^{iz}$ so that
$$x-x^{-1}=2ik\pi \implies x^2-2ik\pi x-1=0\;,\;\;k\in\Bbb Z$$
which by the quadratic formula forms
$$x=ik\pi \pm\sqrt{1-k^2\pi^2} \implies e^{iz}=ik\pi \pm\sqrt{1-k^2\pi^2}\;,\;\;k\in\Bbb Z$$
from which we can take logarithms of both sides to solve for $z$.